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If you're having PETSc use coloring and have confirmed that the stencil is sufficient, then it would be nonsmoothness (again, consider the limiter you've chosen) preventing quadratic convergence (assuming that doesn't kick in eventually). Note that assembling a Jacobian of a second order TVD operator requires at least second neighbors while the first order needs only first neighbors, thus is much sparser and needs fewer colors to compute. I expect you're either not exploiting that in the timings or something else is amiss. You can run with `-log_view -snes_view -ksp_converged_reason` to get a bit more information about what's happening.
"Zou, Ling via petsc-users" <petsc-users@mcs.anl.gov> writes:
> Barry, thank you.
> I am not sure if I exactly follow you on this:
> “Are you forming the Jacobian for the first and second order cases inside of Newton?”
>
> The problem that we deal with, heat/mass transfer in heterogeneous systems (reactor system), is generally small in terms of size, i.e., # of DOFs (several k to maybe 100k level), so for now, I completely rely on PETSc to compute Jacobian, i.e., finite-differencing.
>
> That’s a good suggestion to see the time spent during various events.
> What motivated me to try the options are the following observations.
>
> 2nd order FVM:
>
> Time Step 149, time = 13229.7, dt = 100
>
> NL Step = 0, fnorm = 7.80968E-03
>
> NL Step = 1, fnorm = 7.65731E-03
>
> NL Step = 2, fnorm = 6.85034E-03
>
> NL Step = 3, fnorm = 6.11873E-03
>
> NL Step = 4, fnorm = 1.57347E-03
>
> NL Step = 5, fnorm = 9.03536E-04
>
> Solve Converged!
>
> 1st order FVM:
>
> Time Step 149, time = 13229.7, dt = 100
>
> NL Step = 0, fnorm = 7.90072E-03
>
> NL Step = 1, fnorm = 2.01919E-04
>
> NL Step = 2, fnorm = 1.06960E-05
>
> NL Step = 3, fnorm = 2.41683E-09
>
> Solve Converged!
>
> Notice the obvious ‘stagnant’ in residual for the 2nd order method while not in the 1st order.
> For the same problem, the wall time is 10 sec vs 6 sec. I would be happy if I can reduce 2 sec for the 2nd order method.
>
> -Ling
>
> From: Barry Smith <bsm...@petsc.dev>
> Date: Sunday, March 3, 2024 at 12:06 PM
> To: Zou, Ling <l...@anl.gov>
> Cc: petsc-users@mcs.anl.gov <petsc-users@mcs.anl.gov>
> Subject: Re: [petsc-users] 'Preconditioning' with lower-order method
> Are you forming the Jacobian for the first and second order cases inside of Newton? You can run both with -log_view to see how much time is spent in the various events (compute function, compute Jacobian, linear solve, .. . ) for the two cases
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>
> Are you forming the Jacobian for the first and second order cases inside of Newton?
>
> You can run both with -log_view to see how much time is spent in the various events (compute function, compute Jacobian, linear solve, ...) for the two cases and compare them.
>
>
>
>
> On Mar 3, 2024, at 11:42 AM, Zou, Ling via petsc-users <petsc-users@mcs.anl.gov> wrote:
>
> Original email may have been sent to the incorrect place.
> See below.
>
> -Ling
>
> From: Zou, Ling <l...@anl.gov<mailto:l...@anl.gov>>
> Date: Sunday, March 3, 2024 at 10:34 AM
> To: petsc-users <petsc-users-boun...@mcs.anl.gov<mailto:petsc-users-boun...@mcs.anl.gov>>
> Subject: 'Preconditioning' with lower-order method
> Hi all,
>
> I am solving a PDE system over a spatial domain. Numerical methods are:
>
> * Finite volume method (both 1st and 2nd order implemented)
> * BDF1 and BDF2 for time integration.
> What I have noticed is that 1st order FVM converges much faster than 2nd order FVM, regardless the time integration scheme. Well, not surprising since 2nd order FVM introduces additional non-linearity.
>
> I’m thinking about two possible ways to speed up 2nd order FVM, and would like to get some thoughts or community knowledge before jumping into code implementation.
>
> Say, let the 2nd order FVM residual function be F2(x) = 0; and the 1st order FVM residual function be F1(x) = 0.
>
> 1. Option – 1, multi-step for each time step
> Step 1: solving F1(x) = 0 to obtain a temporary solution x1
> Step 2: feed x1 as an initial guess to solve F2(x) = 0 to obtain the final solution.
> [Not sure if gain any saving at all]
>
>
> 1. Option -2, dynamically changing residual function F(x)
> In pseudo code, would be something like.
>
> snesFormFunction(SNES snes, Vec u, Vec f, void *)
> {
> if (snes.nl_it_no < 4) // 4 being arbitrary here
> f = F1(u);
> else
> f = F2(u);
> }
>
> I know this might be a bit crazy since it may crash after switching residual function, still, any thoughts?
>
> Best,
>
> -Ling
